Question

in a triangle, if the square on one side is equal to the sum of the squares on the other two sides, then provethat the angles opposite to the first side is a right angle.

Answer 1

Hey there !! 


→ Given :- A ∆ABC in which AC² = AB² + BC².

→ To prove :- $\angle B$  = 90° .

→ Construction :- Draw a ∆DEF such that DE = AB, EF = BC and $\angle E$ = 90°.

→ Proof :-)


▶ In ∆DEF , we have $\angle E$ = 90°.

So, by Pythagoras' theorem , we have

=> DF² = DE² + EF².

=> DF² = AB² + BC². ........(1).
[ => DE = AB and EF = BC. ]


But,
AC² = AB² + BC². ......... (2). [Given]


➡ From equation (1) and (2), we get 

=> AC² = DF².
[ By taking under root both side, we get ]

=> AC = DF.


▶Now, in ∆ABC and ∆DEF, we have

=> AB = DE. [By construction]

=> BC = EF. [By construction]

=> AC = DF. [ By showed above]

→ By SSS congruency criteria,

$\bf{ \therefore \triangle ABC \cong \triangle DEF . }$

→ Hence, $\huge \boxed{ \boxed{ \bf \angle B = \angle E = 90 \degree }}$


✔✔ Hence, it is proved ✅✅.

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